if posssible, could you only give me a few theorems in order to assist me in this question. Thankyou in advance! Links to simple websites would also be appreciated.
In triangle $ABC$ $F$ is midpoint of $AC$ $E$ is midpoint of $AB$
$X$ and $Y$ are points along $CB$
If $EX$ and $FY$ are parallel and $EFYX$ is a parallelogram, what is the ratio of the area of $EFYX$ to $ABC$?
All I understand is that similarity or congruency should somehow be used.
On
Stewart's Theorem and the Law of Cosines are useful for this type of problem. Here is a helpful link:
http://www.artofproblemsolving.com/Wiki/index.php/Stewart's_Theorem
Also Menelaus' Theorem will be helpful:
http://en.wikipedia.org/wiki/Menelaus'_theorem
Good luck!
Show/prove/use the following:
$\;FE\;$ is a midsegment of $\;\Delta ABC\;$ and thus $\;\begin{cases}(1)&FE//BC\\{}\\(2)&|FE|=\frac12|BC|\end{cases}\;$
Not only that: the midsegment $\;FE\;$ bisects any segment from vertex $\;A\;$ to side $\;BC\;$ , so:
$$Area_{EFYX}=\frac{|EF|\cdot h}2\;\;,\;\;h=\text{height of paralellogram}\;EFYX$$
$$Area_{\Delta ABC}=\frac{BC\cdot H}2\;\;,\;\;H=\text{height to $\;BC\;$ in triangle}\;\;\Delta ABC$$
But, as mentioned above, $\;h=\frac H2\;$ , so...